Simple enough,
Just consider Newton's second law of dynamics:
$$
\frac{d\mathbf{p}}{dt} = \sum_i \mathbf{{F_i}} = \mathbf{F}
$$
where bold characters represent vectors and $\mathbf{F}$ is just the total resulting force.
In order to give a definition of the kinetic energy, let's just consider an ordinary system that will enable us to build some intuition. Here we'll consider an object of constant mass $m$ and velocity $\mathbf{v}$. Then by definition of the momentum $\mathbf{p}$:
$$
\mathbf{p} = m \mathbf{v}.
$$
Now let's insert this in Newton's second law and perform a dot product with $\mathbf{v}$ on both sides with the idea to latter integrate the formula :
$$
\begin{eqnarray}
\mathbf{v} \cdot \frac{d (m ~\mathbf{v})}{dt} &=& \mathbf{v} \cdot \mathbf{F}\\
\frac{d }{dt} \left(\frac{m ~\mathbf{v}^2}{2}\right) &=& \mathbf{v} \cdot \mathbf{F}\\
d \left(\frac{m ~\mathbf{v}^2}{2}\right) &=& \mathbf{v} \cdot \mathbf{F} ~ dt
\end{eqnarray}
$$
Then we can already see that the natural quantity $K = \frac{m ~\mathbf{v}^2}{2}$ varies if the quantity $ \mathbf{v} \cdot \mathbf{F} ~ dt$ varies. We then can decide to call this new quantity $K$ the kinetic energy since it is a new quantity that depends on the speed only and we can refine the expression of $ \mathbf{v} \cdot \mathbf{F} ~ dt$ :
$$
\frac{d\mathbf{s}}{dt} \cdot \mathbf{F} ~ dt = \mathbf{F} \cdot d\mathbf{s}
$$
where we used that $\mathbf{v} = \frac{d\mathbf{s}}{dt}$.
As we did for $K$ we'll name this new quantity $W$. We'll call it work since it represents an amount of force carried along some path. Work properly defined is the integral of this quantity between two points A and B.
$$ W_{AB} = \int_A^B \mathbf{F} \cdot d\mathbf{s} $$
So this should settle the matter.
One last point, this little demonstration was meant to show how the $\frac{1}{2}$ factor naturally arrises from the fundamental equations of dynamics. However, nothing forced us except some good sense, to chose $K = \frac{m ~\mathbf{v}^2}{2}$ as the definition of the kinetic energy and it has, now, been accepted as an international convention or, let's just say, usage. Therefore, anyone giving the amount of kinetic energy would use this relation for the kinetic energy -- otherwise, this person would have to specify it before hand, and faces a cumbersome process that may loose this person's audience.
Remark : In the derivation, we didn't use the condition that the mass was constant, it never left the differential operator $\frac{d}{dt}$.
